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Q.Balance the following redox reaction by ion - electron method in basic medium.
MnO4^-(aq) + Br^-(aq) --> MnO2(s) + BrO3^-(aq)

Andhra Pradesh BieapBIEAP Intermediate Board (1st Year) 2020Subjective· 4mImportance★★★★★
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The balanced ionic equation is 2MnO4^-(aq) + Br^-(aq) + H2O(l) -> 2MnO2(s) + BrO3^-(aq) + 2OH^-(aq), obtained by combining the reduction and oxidation half-reactions.

Step 1 — Identify the two half-reactions

Reduction: MnO4^- -> MnO2 (Mn goes from +7 to +4, a gain of 3 electrons)

Oxidation: Br^- -> BrO3^- (Br goes from -1 to +5, a loss of 6 electrons)

Step 2 — Balance each half-reaction (basic medium: balance O with H2O, then H with OH-)

Reduction half:

MnO4^- + 2H2O + 3e^- -> MnO2 + 4OH^-

(O: 4+2=6 both sides; H: 4 both sides; charge: -1-3=-4 both sides)

Oxidation half:

Br^- + 6OH^- -> BrO3^- + 3H2O + 6e^-

(O: 6 both sides; H: 6 both sides; charge: -7 both sides)

Step 3 — Equalise electrons and add

Multiply the reduction half by 2 (to get 6 electrons, matching the oxidation half):

2MnO4^- + 4H2O + 6e^- -> 2MnO2 + 8OH^-

Add to the oxidation half:

2MnO4^- + 4H2O + Br^- + 6OH^- + 6e^- -> 2MnO2 + 8OH^- + BrO3^- + 3H2O + 6e^-

Step 4 — Cancel common terms

Cancel 6e^- from both sides. …

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